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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>2. Solve the following system</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
\begin{cases}
\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2},\quad 0&lt;x&lt;2, ~t&gt;0,\\
u(x, 0)=3 \sin (2 \pi x),\quad 0&lt;x&lt;2,\\
u(0, t)=0,\quad u(2, t)=0,\quad t&gt;0.
\end{cases}
\end{equation*}
</div>
<p class="continuation">(a) Take the Laplace transform and apply the initial condition</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
\frac{\mathrm{d}^2 U(x, s)}{\mathrm{d} x^2}=s U(x, s)-u(x, 0)=s U(x, s)-3 \sin (2 \pi x).
\end{equation*}
</div>
<p class="continuation">We write this equation as a non-homogeneous, second order linear constant coefficient equation</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation}
\frac{\mathrm{d}^2 U(x, s)}{\mathrm{d} x^2}-s U(x, s)=-3 \sin (2 \pi x).\tag{8.6.2}
\end{equation}
</div>
<p class="continuation">(b)The general solution of (<a href="" class="xref" data-knowl="./knowl/Eq9_1.html" title="Equation 8.6.2">(8.6.2)</a>) can be obtained by applying the method we have learned in ODE and we get</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
U(x, s)=C_1 e^{\sqrt{s} x}+C_2 e^{-\sqrt{s}x}+\frac{3}{(s+ 4\pi^2)} \sin (2 \pi x).
\end{equation*}
</div>
<p class="continuation">We note the Laplace transform of the boundary conditions give</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
u(0, t)=0 ~\to~ U(0, s)=0,\qquad u(2, t)=0 ~\to~ U(2, s)=0.
\end{equation*}
</div>
<p class="continuation">So we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
0=U(0, s)=C_1+C_2,\quad 0=U(2, s)=C_1 e^{2\sqrt{s} }+C_2 e^{-2\sqrt{s}}
\end{equation*}
</div>
<p class="continuation">which gives <span class="process-math">\(C_1=0, C_2=0\)</span> and we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
U(x, s)=\frac{3}{(s+4 \pi^2)} \sin (2 \pi x).
\end{equation*}
</div>
<p class="continuation">(c) From the table, we find that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
{\mathcal L}^{-1}[U(x, s)]=3 e^{-4 \pi^2 t} \sin (2 \pi x).
\end{equation*}
</div>
<p class="continuation">Thus</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq9_1.html">
\begin{equation*}
u(x, t)=3 e^{-4 \pi^2 t} \sin (2 \pi x).
\end{equation*}
</div>
<span class="incontext"><a href="sec8_6.html#p-504" class="internal">in-context</a></span>
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